ios - 在不支持并发的函数中调用'async' swift ios Xcode async/await

Question

我正在尝试将 async/await 与 Swift 5.5 一起使用。我有我的异步函数，但每当我尝试调用它时，我都会收到此错误:

'async' call in a function that does not support concurrency

这是代码示例:

class TryThis {
    func getSomethingLater(_ number: Double) async -> String {
        // test - sleep for 3 seconds, then return
        Thread.sleep(forTimeInterval: 3)
        return String(format: ">>>%8.2f<<<", number)
    }
}

let tryThis = TryThis()

let result = await tryThis.getSomethingLater(3.141592653589793238462)
print("result: \(result)")

这个有什么解决办法？？

Answer 1

这里的答案是“await getSomethingLater”必须从异步上下文中调用。从字面上看，这意味着改变这一点:

let result = await tryThis.getSomethingLater(3.141592653589793238462)
print("result: \(result)")

进入这个:

Task {
    let result = await tryThis.getSomethingLater(3.141592653589793238462)
    print("result: \(result)")
}

所以整个事情变成了:

class TryThis {
    func getSomethingLater(_ number: Double) async -> String {
        // test - sleep for 3 seconds, then return
        Thread.sleep(forTimeInterval: 3)
        return String(format: ">>>%8.2f<<<", number)
    }
}

let tryThis = TryThis()

Task {
    let result = await tryThis.getSomethingLater(3.141592653589793238462)
    print("result: \(result)")
}

这是输出:

result: >>> 3.14<<<

Meet async/await in Swift 中有大量信息来自 WWDC21 的视频。